Calculate the molality of each of the following solutions: (a) 14.3 g of sucrose (C12H22O11) in 685 g of water,
(b) 7.15 moles of ethylene glycol (C2H6O2) in 3505 g of water.

(a) The molality of this solution is 0.0613

(b)The molality of this solution is 2.04

Explanation:

The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.

The Molality of a solution is determined by the expression:

Molality is expressed in units

(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 , then 14.3 grams of the compound represents the following number of moles:

0.042 moles

Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:

Solving:

molality= 0.0613

The molality of this solution is 0.0613

(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:

Solving:

molality= 2.04

The molality of this solution is 2.04